如何实现树的接口

Question

我有一个树结构TableOfContents：

export interface TableOfContents {
  id: number;
  title: string;
  isExpanded: boolean;
  children: TableOfContents[];
}

我需要一个函数，用于搜索树数组中的项，并希望使其成为通用的。所以我做了下一个：

export interface TreeNode {
  [key: string]: any;
  children?: TreeNode[];
}

export interface TableOfContents extends TreeNode {
  id: number;
  title: string;
  isExpanded: boolean;
}

export const findInTrees = <TreeType extends TreeNode>(
  trees: TreeType[],
  callback: (el: TreeType) => boolean
): TreeType | null => {
  const tree = trees.find(tree => callback(tree));
  if (tree) {
    return tree;
  }
  for (const tree of trees) {
    if (tree.children) {
      const result = findInTrees(tree.children, callback);
      if (result) {
        return result;
      }
    }
  }
  return null;
};

但是在行TS2345: Argument of type 'TreeNode[]' is not assignable to parameter of type 'TreeType[]'.上有一个错误const result = findInTrees(tree.children, callback);

你能帮我找个解决办法吗？

Answer 1

在您的TreeNode.接口中，您的代码说children应该是实现TreeNode的任何东西的数组，而实际上您的意思是children应该是相同类型的TreeNode的数组。因此，使用this类型：

export interface TreeNode {
  [key: string]: any;
  children?: this[];
}

Answer 2

export type INode<T = any> = T & {
    nodes: Array<INode<T>>;
}



const family: INode<{name?: string}> = {
    name: 'Bogdan',
    nodes: []
};